Determine all $f:\mathbb{R}\to\mathbb{R}$ such that $(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)$.

Question

Determine all $f:\mathbb{R}\to\mathbb{R}$ such that $$(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)\,.$$

Help pls! Thank you.

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Edit (by Batominovski).

See the edit history of this question. Originally, the functional equation $$(x+y)\,\big(f(x){\color{red}-}f(y)\big)=f(x^2)-f(y^2)$$ was in the title, but the body asked about $$(x+y)\,\big(f(x){\color{red}+}f(y)\big)=f(x^2)-f(y^2).$$ The OP has not clarified which functional equation is the real question.

Note : The functional equation that Mr. Batominovski solved first was the real question. I’m sorry for mistake in typing. Thank you for helping!

Answer 1

Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $$(x+y)\,\big(f(x)-f(y)\big)=f(x^2)-f(y^2)\,.$$ for all $x,y\in\mathbb{R}$. Let $c:=f(0)$. Plugging $y:=0$ into the original functional equation yields $$f(x^2)=x\,\big(f(x)-c\big)$$ for each $x\in\mathbb{R}$. Using this result in the original functional equation, we obtain $$(x+y)\,\big(f(x)-f(y)\big)=x\,\big(f(x)-c\big)-y\,\big(f(y)-c\big)$$ for all $x,y\in\mathbb{R}$. After some algebraic manipulations, we get $$y\,\big(f(x)-c\big)=x\,\big(f(y)-c\big)$$ for all $x,y\in\mathbb{R}$. Hence, for any real numbers $x,y\neq 0$, $$\frac{f(x)-c}{x}=\frac{f(y)-c}{y}\,.$$ Thus, there exists a constant $k$ such that $$\frac{f(x)-c}{x}=k$$ for every real number $x\neq 0$. This shows that $$f(x)=kx+c\text{ for all real numbers $x\neq 0$}\,.$$ However, $f(0)=c$ implies that $$f(x)=kx+c\text{ for all $x\in\mathbb{R}$}\,.$$ It is easy to see that any function of the form above (i.e., $k$ and $c$ can be arbitrary real numbers) is a solution.

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Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $$(x+y)\,\big(f(x)+f(y)\big)=f(x^2)-f(y^2)$$ for all $x,y\in\mathbb{R}$. Then, by letting $y:=x$ in the original functional equation, we obtain $$4x\,f(x)=0$$ for all $x\in\mathbb{R}$. In particular, this shows that $f(x)=0$ for every $x\in\mathbb{R}_{\neq 0}$. Now, by letting $x:=1$ and $y:=0$ in the original functional equation and using $f(1)=0$, we have $$f(0)=-f(0)\,.$$ Therefore, $f(0)=0$. Ergo, $$f(x)=0\text{ for all }x\in\mathbb{R}\,,$$ which is easily seen to be a solution.

Edit. The solution for another functional equation was added, because Tavish remarked that the functional equation was changed. The OP wrote two different functional equations. When I edited the question to fix grammatical mistakes and to improve the format, I inadvertently changed one functional equation the OP had written to another. The OP has not yet clarified which functional equation is the real question.

Answer 2

From

$$ \cases{ (x-y)(f(x)-f(-y))=f(x^2)-f(y^2)\\ -(x-y)(f(-x)-f(y))=f(x^2)-f(y^2) }\Rightarrow f(x)+f(-x)=f(y)+f(-y)\Rightarrow f(-x)+f(x) = c $$

now

$$ \cases{ x+y=\frac{f(x^2)-f(y^2)}{f(x)-f(y)}\\ x-y=\frac{f(x^2)-f(y^2)}{f(x)+f(y)-c} }\Rightarrow \cases{ f(x) = \frac c2+\frac{f(x^2)-f(y^2)}{x^2-y^2}x\\ f(y) = \frac c2+\frac{f(x^2)-f(y^2)}{x^2-y^2}y }\Rightarrow f(x) = k x+c_0 $$